Chemistry · Lesson
Reaction ratios and stoichiometric calculations
Use equation coefficients as amount ratios, then convert between mass and amount at the appropriate ends of a calculation.
Back to the Chemistry mapBy the end, you can
- Read a mole ratio from a balanced equation.
- Follow a mass-to-amount-to-product calculation.
- State the assumptions behind a predicted product amount.
Read a proportional relationship
For 2 H₂ + O₂ → 2 H₂O, consumption of 2 mol H₂ corresponds to consumption of 1 mol O₂ and formation of 2 mol H₂O in the stated net reaction. The same relation holds for suitable multiples. The coefficients express relative counts or amounts of the specified species.
They do not express gram ratios. Two grams of H₂ do not simply combine with one gram of O₂ to make two grams of water. Different substances have different molar masses. Convert masses to amounts before applying the coefficient ratio, then convert the desired amount back to mass if required.
Put each operation where it belongs
A mass-to-mass calculation follows four stages: starting-substance mass → its amount → desired-product amount → desired-product mass. Divide by the starting substance’s molar mass, multiply by the balanced coefficient ratio, then multiply by the product’s molar mass. The units show which conversion is happening.
For an arithmetic example, 0.400 mol H₂ corresponds to 0.400 mol H₂O because the coefficient ratio is 2/2. With M(H₂O) = 18.016 g/mol, that product amount corresponds to 7.2064 g before final rounding. The equality of these mole amounts is specific to this equation ratio, not a rule for every reactant and product.
| Stage | Operation | Result |
|---|---|---|
| H₂ mass to amount | 0.8064 g ÷ 2.016 g/mol | 0.4000 mol H₂ |
| Equation ratio | 0.4000 mol H₂ × (2 mol H₂O / 2 mol H₂) | 0.4000 mol H₂O |
| Product amount to mass | 0.4000 mol H₂O × 18.016 g/mol | 7.2064 g H₂O |
Theoretical water calculation using supplied molar masses and a sufficient oxygen amount. Arithmetic values are shown before measurement-specific rounding.
A theoretical result needs its conditions
To infer a product amount from just one starting reactant, assume the other needed reactant is sufficient, the stated reaction accounts for the conversion, and the chosen reactant converts as specified. Otherwise the one-reactant prediction may exceed what the available mixture can make. The limiting-reactant lesson compares all required inputs.
The calculation is an upper accounting prediction under those assumptions. It is not a practical procedure for combining hydrogen and oxygen, nor evidence that a real reaction has occurred. Actual conversion, side reactions and product recovery need measurements. State those boundaries alongside the numerical result.
Check the direction and magnitude
After calculating, reread the coefficient ratio. In N₂ + 3 H₂ → 2 NH₃, 1 mol N₂ can produce 2 mol NH₃ under complete-conversion and excess-hydrogen assumptions. The count of molecules can increase while every element count remains conserved. Conservation does not require the total number of molecules to stay the same.
For mass calculations, verify that the total participating mass is accounted for within the stated material boundary. A product can weigh more than the selected reactant because another reactant contributes atoms and mass. A quick ratio or unit check helps detect mistakes, but it does not replace examining reaction assumptions.
Worked examples
Apply a coefficient ratio
Assuming sufficient hydrogen and complete conversion by N₂ + 3 H₂ → 2 NH₃, what ammonia amount corresponds to 0.250 mol N₂?
- The balanced ratio is 2 mol NH₃ per 1 mol N₂.
- Multiply 0.250 mol N₂ by 2 mol NH₃ / 1 mol N₂.
- Check nitrogen atoms: 0.250 mol N₂ contains 0.500 mol N atoms; the predicted NH₃ contains the same amount.
0.500 mol NH₃ under the stated assumptions. Hydrogen consumption would be 0.750 mol.
Try it yourself
For the balanced water equation, how much O₂ is needed for 0.400 mol H₂?
Show the answer and reasoning
0.200 mol O₂.
Use the O₂:H₂ coefficient ratio 1:2.
Do equation coefficients directly give gram ratios?
Show the answer and reasoning
No.
They give species-count or amount ratios. Molar masses connect those amounts to grams.
Under the stated complete reaction, 0.600 mol H₂ can form how much NH₃ if nitrogen is sufficient?
Show the answer and reasoning
0.400 mol NH₃.
Multiply by the NH₃:H₂ coefficient ratio 2/3.
Can the product mass exceed the mass of one selected reactant without violating mass accounting?
Show the answer and reasoning
Yes.
Other reactants can contribute material. Check the whole participating system.
What to carry forward
- Equation coefficients give species amount ratios.
- Use mass → amount → coefficient ratio → amount → mass.
- Retain sufficient-reactant and conversion assumptions.
- Conserve elements and total participating mass, not necessarily molecule number.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.